Chapter 2 Kinematics in One Dimension
Chapter 2
KINEMATICS IN ONE DIMENSION
PREVIEW
Kinematics is the study of how things move – how far (distance and displacement), how
fast (speed and velocity), and how fast that how fast changes (acceleration). We say that
an object moving in a straight line is moving in one dimension, and an object which is
moving in a curved path (like a projectile) is moving in two dimensions. We relate all
these quantities with a set of equations called the kinematic equations.
The content contained in all sections of chapter 2 of the textbook is included on the AP
Physics B exam.
QUICK REFERENCE
Important Terms
acceleration
the rate of change in velocity
acceleration due to gravity
the acceleration of a freely falling object in the absence of air resistance, which
near the earth’s surface is approximately 10 m/s2
acceleration-time graph
plot of the acceleration of an object as a function of time
average acceleration
the acceleration of an object measured over a time interval
average velocity
the velocity of an object measured over a time interval; the displacement of an
object divided by the change in time during the motion
constant (or uniform) acceleration
acceleration which does not change during a time interval
constant (or uniform) velocity
velocity which does not change during a time interval
displacement
change in position in a particular direction (vector)
distance
the length moved between two points (scalar)
free fall
motion under the influence of gravity
initial velocity
the velocity at which an object starts at the beginning of a time interval
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Chapter 2 Kinematics in One Dimension
instantaneous
the value of a quantity at a particular instant of time, such as instantaneous
position, velocity, or acceleration
kinematics
the study of how motion occurs, including distance, displacement, speed, velocity,
acceleration, and time
position-time graph
the graph of the motion of an object that shows how its position varies with time
speed
the ratio of distance to time
velocity
ratio of the displacement of an object to a time interval
velocity-time graph
plot of the velocity of an object as a function of time, the slope of which is
acceleration, and the area under which is displacement
Equations and Symbols
v  vo
t
v  v o  at
a
1
(v o  v)t
2
1
x  v o t  at 2
2
2
2
v  v o  2 a x
x 
where
Δx = displacement (final position – initial
position)
v = velocity or speed at any time
vo = initial velocity or speed
t = time
a = acceleration
Ten Homework Problems
Chapter 2 Problems 2, 10, 13, 30, 40, 43, 58, 59, 73, 78
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Chapter 2 Kinematics in One Dimension
DISCUSSION OF SELECTED SECTIONS
2.1 Displacement
Distance d can be defined as total length moved. If you run around a circular track, you
have covered a distance equal to the circumference of the track. Distance is a scalar,
which means it has no direction associated with it. Displacement Δx, however, is a
vector. Displacement is defined as the straight-line distance between two points, and is a
vector which points from an object’s initial position xo toward its final position xf. In our
previous example, if you run around a circular track and end up at the same place you
started, your displacement is zero, since there is no distance between your starting point
and your ending point. Displacement is often written in its scalar form as simply Δx or x.
2.2 Speed and Velocity
Average speed is defined as the amount of distance a moving object covers divided by
the amount of time it takes to cover that distance:
average speed = v 
distance
d

elapsed time
t
where v stands for speed, d is for distance, and t is time.
Average velocity is defined a little differently than average speed. While average speed is
the total change in distance divided by the total change in time, average velocity is the
displacement divided by the change in time. Since velocity is a vector, we must define it
in terms of another vector, displacement. Oftentimes average speed and average velocity
are interchangeable for the purposes of the AP Physics B exam. Speed is the magnitude
of velocity, that is, speed is a scalar and velocity is a vector. For example, if you are
driving west at 50 miles per hour, we say that your speed is 50 mph, and your velocity is
50 mph west. We will use the letter v for both speed and velocity in our calculations, and
will take the direction of velocity into account when necessary.
2.3 Acceleration
Acceleration tells us how fast velocity is changing. For example, if you start from rest on
the goal line of a football field, and begin walking up to a speed of 1 m/s for the first
second, then up to 2 m/s for the second second, then up to 3 m/s for the third second, you
are speeding up with an average acceleration of 1 m/s for each second you are walking.
We write
a
v 1 m / s
m

 1m / s / s  1 2
t
1s
s
In other words, you are changing your speed by 1 m/s for each second you walk. If you
start with a high velocity and slow down, you are still accelerating, but your acceleration
would be considered negative, compared to the positive acceleration discussed above.
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Chapter 2 Kinematics in One Dimension
Usually, the change in speed v is calculated by the final speed vf minus the initial speed
vo. The initial and final speeds are called instantaneous speeds, since they each occur at a
particular instant in time and are not average speeds.
2.5 Applications of the Equations of Kinematics for Constant
Acceleration
Kinematics is the study of the relationships between distance and displacement, speed and
velocity, acceleration, and time. The kinematic equations are the equations of motion
which relate these quantities to each other. These equations assume that the acceleration
of an object is uniform, that is, constant for the time interval we are interested in. The
kinematic equations listed below would not work for calculating velocities and
displacements for an object which is accelerating erratically. Fortunately, the AP Physics
B exam generally deals with uniform acceleration, so the kinematic equations listed
above will be very helpful in solving problems on the test.
2.6 Freely Falling Bodies
An object is in free fall if it is falling freely under the influence of gravity. Any object,
regardless of its mass, falls near the surface of the Earth with an acceleration of 9.8 m/s2,
which we will denote with the letter g. We will round the free fall acceleration g to 10
m/s2 for the purpose of the AP Physics B exam. This free fall acceleration assumes that
there is no air resistance to impede the motion of the falling object, and this is a safe
assumption on the AP Physics B test unless you are told differently for a particular
question on the exam.
Since the free fall acceleration is constant, we may use the kinematic equations to solve
problems involving free fall. We simply need to replace the acceleration a with the
specific free fall acceleration g in each equation. Remember, anytime velocity and
acceleration are in opposite directions (as when a ball is rising after being thrown
upward), you must give one of them a negative sign.
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Chapter 2 Kinematics in One Dimension
1 2
at
2
1
1 m  13 m / s 2 t 2
2
t  0.4 s
y


Example 1 A girl is holding a ball as she steps onto a tall elevator
on the ground floor of a building. The girl holds the ball at a height
of 1 meter above the elevator floor. The elevator begins
accelerating upward from rest at 3 m/s2. After the elevator
accelerates for 5 seconds, find
(a) the speed of the elevator
(b) the height of the floor of the
elevator above the ground.
At the end of 5 s, the girl lets go of the ball from a height of 1 meter above the floor of
the elevator. If the elevator continues to accelerate upward at 3 m/s2, describe the motion
of the ball
(c) relative to the girl’s hand,
(d) relative to the ground.
(e) Determine the time after the ball is released that it will make contact with the floor.
(f) What is the height above the ground of the ball and floor when they first make
contact?
Solution:
(a) v  vo  at  0  3 m / s 2 5s   15 m / s upward
1
2
(b) y  v o t  at 2  0 


1
3.0 m / s 2 5.0 s 2  37 .50 m
2
(c) When the girl releases the ball, both she and the ball are moving with a speed of 15
m/s upward. However, the girl continues to accelerate upward at 3 m/s2, but the ball
ceases to accelerate upward, and the ball’s acceleration is directed downward at g = 10
m/s2, that is, it is in free fall with an initial upward velocity of 15 m/s. Therefore the ball
will appear to the girl to fall downward with an acceleration of 3 m/s2 – (- 10 m/s2) = 13
m/s2 downward, and will quickly fall below her hand.
(d) Someone watching the ball from the ground would simply see the ball rising upward
with an initial velocity of 15 m/s, and would watch it rise to a maximum height, at which
point it would be instantaneously at rest (provided it doesn’t strike the floor of the
elevator before it reaches its maximum height).
(e) When the ball is released, it is traveling upward with a speed of 15 m/s, has a
downward acceleration of 13 m/s2 relative to the floor, and is at a height y = 1 m above
the floor. The time it takes to “fall” to the floor is
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Chapter 2 Kinematics in One Dimension
(f) In the total time of 0.4 s, the elevator floor has moved up a distance of
y 
1
1
a t 2   3 m / s 2 0.4s 2  0.24 m

2 e
2
Thus, the ball and elevator floor collide at a height above the ground of
37.50 m + 0.24 m = 37.74 m
2.7 Graphical Analysis of Velocity and Acceleration
Let’s take some time to review how we interpret the motion of an object when we are
given the information about it in graphical form. On the AP Physics B exam, you will
need to be able to interpret three types of graphs: position vs. time, velocity vs. time, and
acceleration vs. time.
Position vs. time
Consider the position vs. time graph below:
x (m)
x (m)
Δx
P
Δx
Δt
Δt
t (s)
t (s)
x
, and is therefore velocity. The curved graph on
t
the right indicates that the slope is changing. The slope of the curved graph is still
velocity, even though the velocity is changing, indicating the object is accelerating. The
instantaneous velocity at any point on the graph (such as point P) can be found by
drawing a tangent line at the point and finding the slope of the tangent line.
The slope of the graph on the left is
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Chapter 2 Kinematics in One Dimension
Velocity vs. time
Consider the velocity vs. time graph below:
v (m/s)
v (m/s)
Δv
Area
Δt
t (s)
t (s)
v
, and is
t
therefore acceleration. As shown on the figure on the right, the area under a velocity vs.
m
time graph would have units of s   m , and is therefore displacement.
s
As shown in the figure on the left, the slope of a velocity vs. time graph is
Acceleration vs. time
Since the AP Physics B exam generally deals with constant acceleration, any graph of
acceleration vs. time on the exam would likely be a straight horizontal line:
a (m/s2)
a (m/s2)
+5 m/s2
0
0
t(s)
-5 m/s2
t(s)
This graph on the left tells us that the acceleration of this object is positive. If the object
were accelerating negatively, the horizontal line would be below the time axis, as shown
in the graph on the right.
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Chapter 2 Kinematics in One Dimension
Example 2 Consider the position vs. time graph below representing the motion of a car.
Assume that all accelerations of the car are constant.
H
G
D
x(m)
E
I
J
F
C
B
K
0
t(s)
A
On the axes below, sketch the velocity vs. time and acceleration vs. time graphs for this
car.
v(m/s)
0
t(s)
a(m/s2)
0
t(s)
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Chapter 2 Kinematics in One Dimension
Solution:
The car starts out at a distance behind our reference point of zero, indicated on the graph
as a negative displacement. The velocity (slope) of the car is initially positive and
constant from points A to C, with the car crossing the reference point at B. Between
points C and D, the car goes from a high positive velocity (slope) to a low velocity,
eventually coming to rest (v = 0) at point D. At point E the car accelerates positively from
rest up to a positive constant velocity from points F to G. Then the velocity (slope)
decreases from points G to H, indicating the car is slowing down. It is between these two
points that the car’s velocity is positive, but its acceleration is negative, since the car’s
velocity and acceleration are in opposite directions. The car once again comes to rest at
point H, and then begins gaining a negative velocity (moving backward) from rest at
point I, increasing its speed negatively to a constant negative velocity between points J
and K. At K, the car has returned to its original starting position.
The velocity vs. time graph for this car would look like this:
v(m/s)
B
C
F
G
A
D
E
I
H
0
t(s)
J
K
The acceleration vs. time graph for this car would look like this:
a(m/s2)
E
A
B
C
D
F
G
H
I
J
0
t(s)
27
K
Chapter 2 Kinematics in One Dimension
CHAPTER 2 REVIEW QUESTIONS
For each of the multiple-choice questions below, choose the best answer.
Unless otherwise noted, use g = 10 m/s2 and neglect air resistance.
1. Which of the following statements is
true?
(A) Displacement is a scalar and
distance is a vector.
(B) Displacement is a vector and
distance is a scalar.
(C) Both displacement and distance are
vectors.
(D) Neither displacement nor distance
are vectors.
(E) Displacement and distance are
always equal.
5. A bus starting from a speed of +24
m/s slows to 6 m/s in a time of 3 s. The
average acceleration of the bus is
(A) 2 m/s2
(B) 4 m/s2
(C) 6 m/s2
(D) – 2 m/s2
(E) – 6 m/s2
6. A train accelerates from rest with an
acceleration of 4 m/s2 for a time of 20 s.
What is the train’s speed at the end of 20
s?
(A) 0.25 m/s
(B) 4 m/s
(C) 2.5 m/s
(D) 0.8 m/s
(E) 80 m/s
2. Which of the following is the best
statement for a velocity?
(A) 60 miles per hour
(B) 30 meters per second
(C) 30 km at 45 north of east
(D) 40 km/hr
(E) 50 km/hr southwest
7. A football player starts from rest 10
meters from the goal line and accelerates
away from the goal line at 5 m/s2. How
far from the goal line is the player after 4
s?
(A) 6 m
(B) 30 m
(C) 40 m
(D) 50 m
(E) 60 m
3. A jogger runs 4 km in 0.4 hr, then 8
km in 0.8 hr. What is the average speed
of the jogger?
(A) 10 km/hr
(B) 3 km/hr
(C) 1 km/hr
(D) 0.1 km/hr
(E) 100 km/hr
4. A motorcycle starts from rest and
accelerates to a speed of 20 m/s in a time
of 8 s. What is the motorcycle’s average
acceleration?
(A) 160 m/s2
(B) 80 m/s2
(C) 8 m/s2
(D) 2.5 m/s2
(E) 0.4 m/s2
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Chapter 2 Kinematics in One Dimension
8. A ball is dropped from rest. What is
the acceleration of the ball immediately
after it is dropped?
(A) zero
(B) 5 m/s2
(C) 10 m/s2
(D) 20 m/s2
(E) 30 m/s2
Questions 9 – 11:
A ball is thrown straight upward with a
speed of +12 m/s.
12. Which two of the following pairs of
graphs are equivalent?
(A) x
v
9. What is the ball’s acceleration just
after it is thrown?
(A) zero
(B) 10 m/s2 upward
(C) 10 m/s2 downward
(D) 12 m/s2 upward
(E) 12 m/s2 downward
0
t
(B)
t
x
v
0
t
10. How much time does it take for the
ball to rise to its maximum height?
(A) 24 s
(B) 12 s
(C) 10 s
(D) 2 s
(E) 1.2 s
(C)
x
t
v
0
t
(D)
x
t
v
0
11. What is the approximate maximum
height the ball reaches?
(A) 24 m
(B) 17 m
(C) 12 m
(D) 7 m
(E) 5 m
t
(E)
x
v
t
29
t
0
t
Chapter 2 Kinematics in One Dimension
Questions 13 – 14:
Consider the velocity vs. time graph
below:
13. A which time(s) is the object at rest?
(A) zero
(B) 1 s
(C) 3 s to 4 s
(D) 4 s only
(E) 8 s
14. During which interval is the speed of
the object decreasing?
(A) 0 to 1 s
(B) 1 s to 3 s
(C) 3 s to 4 s
(D) 4 s to 8 s
(E) the speed of the object is never
decreasing in this graph
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Chapter 2 Kinematics in One Dimension
Free Response Question
Directions: Show all work in working the following question. The question is worth 15 points,
and the suggested time for answering the question is about 15 minutes. The parts within a
question may not have equal weight.
1. (15 points)
A cart
on a long horizontal track can move with negligible friction to the left or to the right. During the
time intervals when the cart is accelerating, the acceleration is constant. The acceleration during
other time intervals is also constant, but may have a different value. Data is taken on the motion
of the cart, and recorded in the table below.
displacement
x(m)
2
velocity
v(m/s)
-4
time
t(s)
0
-2
1
-2
2
-2
3
1
6
1
7
0
9
0
10
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Chapter 2 Kinematics in One Dimension
(a) Plot these data points on the v vs. t graph below, and draw the best-fit straight lines between
each data point; that is, connect each data point to the one before it. The acceleration is constant
or zero during each interval listed in the data table.
(b) List all of the times between t = 0 and t = 10 s at which the cart is at rest.
(c) i. During which time interval is the magnitude of the acceleration of the cart the
greatest?
ii. What is the value of this maximum acceleration?
(d) Find the displacement of the cart from x = 0 at a time of 10 s.
(e) On the following graph, sketch the acceleration vs. time graph for the motion of this cart
from t = 0 to t = 10 s.
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Chapter 2 Kinematics in One Dimension
ANSWERS AND EXPLANATIONS TO CHAPTER 2 REVIEW QUESTIONS
Multiple Choice
1. B
Displacement is the straight-line length from an origin to a final position and includes direction,
whereas distance is simply length moved.
2. E
Velocity is a vector and therefore direction should be included.
3. A
Average speed is total distance divided by total time. The total distance covered by the jogger is
12 km and the total time is 1.2 hours, so the average speed is 10 km/hr.
4. D
a
5. E
a
v 20 m / s
m

 2.5 2
t
8s
s
v f  vo
t

6 m / s  24 m / s
m
6 2
3s
s
6. E
v f  vi  at  0  4 m / s 2 20 s   80 m / s


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Chapter 2 Kinematics in One Dimension
7. D
x f  xo  vo t 
1 2
1 m 
2
at  (10 m)  0   5 2 4s   50 m
2
2 s 
8. C
The acceleration due to gravity is 10 m/s2 at all points during the ball’s fall.
9. C
After the ball is thrown, the only acceleration it has is the acceleration due to gravity, 10 m/s2.
10. E
At the ball’s maximum height, vf = 0. Thus,
v f  vo  gt  0
t
12 m / s
 1.2 s
10 m / s 2
11. D
1
1 m 
2
y  gt 2  10 2 1.2 s   7.2 m  7 m
2
2 s 
12. B
Both of these graphs represent motion that begins at a high positive velocity, and slows down to
zero velocity.
13. B
The line crosses the axis (v = 0) at a time of 1 second.
14. A
The object begins with a high negative (backward) velocity at t = 0, then its speed decreases to
zero by a time of 1 s.
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Chapter 2 Kinematics in One Dimension
Free Response Question Solution
(a) 4 points
(b) 2 points
The cart is at rest when the velocity is zero, that is, when the graph crosses the time axis. Thus, v
= 0 at 5 s, 9 s, and 10 s, as well as all points between 9 and 10 s.
(c) i. 1 point
The acceleration can be found by finding the slope of the v vs t graph in a particular interval. The
slope (acceleration) is maximum (steepest) in the time interval from 0 to 1 s.
ii. 2 points
Acceleration = slope of v vs t graph =
 2 m / s   4 m / s 
 2m / s2
1s  0s
(d) 3 points
The displacement of the cart from x = 0 can be found by determining the area under the graph.
Note that the area is negative from 0 to 5 s, and positive from 5 s to 9 s. Don’t forget the initial
displacement of 2 m at t = 0.
Area from 0 to 5 s = 9 squares = - 9 m.
Area from 5 to 10 s = 2.5 squares = +2.5 m
Total displacement from x = 0 is 2 m – 9 m + 2.5 m = - 4.5 m.
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Chapter 2 Kinematics in One Dimension
(e) 3 points
36